Solution :
Given :
Mass attached to the spring = 4 kg
Mass dropped = 6 kg
Force constant = 100 N/m
Initial amplitude = 2 m
Therefore,
a). 

= 10 m/s
Final velocity, v at equilibrium position, v = 5 m/s
Now, 
A' = amplitude = 1.4142 m
b). 
m' = 2m
Hence, 
c). 

Therefore, factor 
Thus, the energy will change half times as the result of the collision.
Answer:
The atmospheric pressure is
.
Explanation:
Given that,
Atmospheric pressure
drop height h'= 27.1 mm
Density of mercury 
We need to calculate the height
Using formula of pressure

Put the value into the formula



We need to calculate the new height




We need to calculate the atmospheric pressure
Using formula of atmospheric pressure

Put the value into the formula


Hence, The atmospheric pressure is
.
Assuming the accleration applied was constant, we have



Then the force applied to the ball is given by


~686newtons on earth and
~1617 newtons on jupiter
the formula is weight = gravitational acceleration * mass of the object